There are two urns. Urn 1 contains 3 white and 2 red ball, urn 2 one white and two
red. First, a ball from urn 1 is randomly chosen and placed into urn 2. Finally, a ball from
urn 2 is picked. This ball be red: What is the probability that the ball transferred from
urn 1 to urn 2 was white?
My answer is 3/5 – 3W over 5 balls in urn 1 as the second event does not tell me anything to influence which ball was transferred since urn 2 already has a red ball anyways. Hope to know if my reasoning is valid!
Best Answer
Method I: (Bayes) There are two scenarios in which a red ball is observed:
I: A white ball is transferred (probability $\frac 35$). In this case, a red ball is observed with probability $\frac 12$. Thus this scenario has probability $$\frac 35 \times \frac 12 = \frac 3{10}$$
II: A red ball is transferred (probability $\frac 25$). Now the probability of drawing a red ball is $\frac 34$ Thus this scenario has probability $$\frac 25\times \frac 34 =\frac 3{10}$$
We see that the two scenarios contribute equally, thus the probability that it was a white ball that was transferred initially is $\boxed {\frac 12}$
Note: as our prior was that the probability the transferred was white was $\frac 35$ we see that the observation of the red ball has caused us to lower our estimate for the probability
Method II (Conditional Probability). A priori, the universe here consists of four possible events: $(W,W),(W,R),(R,R),(R,W)$ according to the color of the transferred ball and the color of the drawn ball. A routine calculation shows that $(W,W),(W,R),(R,R)$ each have probability $\frac 3{10}$ and $(R,W)$ has probability $\frac 1{10}$. We are asked for the probability that the transferred ball is $W$ conditioned on the fact that the drawn ball is $R$ and inspection now shows the answer to be $\frac 12$.